Gauss Law is a general law applying to any closed surface that permits to calculate the field of an enclosed charge by mapping the field on a surface outside the charge distribution. It simplifies the calculation of the electric field with the symmetric geometrical shape of the surface. Its typical case will be the charged particle with spherical symmetry. This article will explain the Gauss law as well as Gauss law formula with examples. Let us learn it!

**What is the Gauss Law?**

Total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of the material. The electric flux through some surface is the electric field multiplied by the surface area, towards perpendicular to the field.

Gauss Law is a general law applicable for any closed surface. It is an important tool as it allows the assessment of the amount of enclosed electric charge. It does the mapping of the field on a surface outside the charge distribution. For geometries having sufficient symmetry, it simplifies the calculation of the electric field.

Source: en.wikipedia.org

Another way of visualizing this is to consider the probe of some areas. That area can measure the electric field perpendicular to it. If it takes any closed surface, then it will obtain a measure of the net electric charge within the surface. It will not depend on how that internal charge is configured.

**The Formula for Gauss Law:**

As per the Gauss theorem, the total charge enclosed in any closed surface is 2proportional to the total flux enclosed by the surface. Therefore, If \phi is the total flux and \epsilon_{0} is the electric constant and the Q is the total electric charge enclosed by the surface is.

Then the formula is: \(\phi = \frac {Q}{ \epsilon_{0} }\)

Where,

\(\phi\) | Electric Flux |

Q | Electric Charge |

\(\epsilon_{0}\) | Permittivity |

It will give the net flux for the surface on the left is non-zero as it encloses a net charge. The net flux for the surface on the right will be zero as it does not enclose any charge. The Gauss law is only a restatement of the Coulombs law for electric charge. If we apply the Gauss theorem to a point charge enclosed by a sphere, you will get back the Coulombâ€™s law easily.

## Solved Examples forÂ Gauss Law Formula

Q.1: If the electric flux throughout a sphere is \(E \times4 \pi r^{2}\). Determine the electric field due to this flux.

Solution: Given parameters are,

\(\phi = E \times4 \pi r^{2}\)

From the formula of the Gauss law we have: \(\phi = \frac {Q}{\epsilon_{0}}\)

Thus, \(E \times 4 \pi r^{2} = \frac {Q}{ \epsilon_{0} }\)

So, \(E = \frac {Q}{(4 \pi r^{2}) \epsilon_{0}}\)

Thus the electric field will be \(\frac {Q}{(4 \pi r^{2}) \epsilon_{0}}.\)

Q.2:Â An electric flux of 2 V-m is going through a sphere in a vacuum space. What will be the charge that origins that flux? Use the Gauss Law Formula.

Solution: From the formula of the Gauss law,

\(\phi = \frac {Q}{\epsilon_{0}}\)

Rearranging it,

\(Q = \phi \times \epsilon_{0}\)

Substituting the values we will get,

\(Q = 2 V-m \times 8.85 \times 10^{-12}\)

\(Q = 17.7 \times 10^{-12} C.\)

Thus charge to generate that flux will be \(17.7 \times 10^{-12} C.\)

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…